Question

Suppose \(A\) and \(B\) are independent events. In expression (1.4.6) we showed that \(A^{c}\) and \(B\) are independent events. Show similarly that the following pairs of events are also independent: (a) \(A\) and \(B^{c}\) and (b) \(A^{c}\) and \(B^{c}\).

Short answer

The provided pairs of events (a) \(A\) and \(B^{c}\), (b) \(A^{c}\) and \(B^{c}\) are indeed independent.

Step-by-step solution

Analyzing Event Pair \(A\) and \(B^{c}\)

To prove that the events \(A\) and \(B^{c}\) are independent, check if their intersection equals the product of their individual probabilities, i.e., does \(P(A \cap B^{c})=P(A)P(B^{c}\)? By using the properties of probabilities, we can express \(P(A \cap B^{c})\) as \(P(A)-P(A \cap B)\). As \(A\) and \(B\) are independent, we have \(P(A \cap B)=P(A)P(B)\), so we can express \(P(A \cap B^{c})\) as \(P(A)-P(A)P(B)\). Hence, \(P(A \cap B^{c})=P(A)(1-P(B))=P(A)P(B^{c})\) since \(B^{c}\) is the complement of \(B\).

Analyzing Event Pair \(A^{c}\) and \(B^{c}\)

To prove that the events \(A^{c}\) and \(B^{c}\) are independent, we need to check if their intersection equals the product of their individual probabilities, i.e., does \(P(A^{c} \cap B^{c})=P(A^{c})P(B^{c})\)? By using the properties of probabilities, we can express \(P(A^{c} \cap B^{c})\) as \(1-P(A \cup B)\). Now \(P(A \cup B)=P(A)+P(B)-P(A \cap B)\), and we know that \(A\) and \(B\) are independent, so \(P(A \cap B)=P(A)P(B)\), thus we can express \(P(A^{c} \cap B^{c})\) as \(1-P(A)-P(B)+P(A)P(B) = (1-P(A))(1-P(B))=P(A^{c})P(B^{c})\) since \(A^{c}\) and \(B^{c}\) are the complements of \(A\) and \(B\) respectively.